Prove $\lim_{n\rightarrow \infty} n^{\frac{1}{n}} = 1$ Prove that $$\lim_{n\rightarrow \infty} n^{\frac{1}{n}} = 1$$
Hint: write $$n^{\frac{1}{n}} = 1 + \epsilon_n, 0 < \epsilon_n $$ then $$n = (1+\epsilon_n)^n > 1 + \frac{n(n+1)}{2}\epsilon_n^2$$ by the binomial theorem.
I am not sure how to establish this result. The only manipulation I see is taking $\frac{1}{n}$ on both sides of the inequality in the hint, but I do not see how that can help me establish a result.
 A: If 
$
n^{\frac{1}{n}} = 1 + \epsilon_n,
$
then by taking $n$th powers, we know that 
$
n = (1 + \epsilon_n)^n
$
and so 
$$
n = (1 + \epsilon_n)^n 
= 1 + \binom{n}{1} \epsilon_n + \binom{n}{2} \epsilon_n ^2 + \cdots 
> 1 + \binom{n}{2} \epsilon_n ^2 
= 1 + \frac{n(n+1)}{2} \epsilon_n ^2 
$$
and therefore.....
$$
0 < \epsilon_n < \sqrt{ \frac{2(n-1)}{n(n+1)} }
$$
But, what happens if we look at the limit as $n \to \infty$? Then, $\epsilon_n \to 0$, and so $n^{\frac{1}{n}} \to 1$.
A: $$ n^{\frac{1}{n}}=e^{ln{n^{\frac{1}{n}}}}=e^{\frac{1}{n}ln(n)}=\\e^{\frac{ln n}{n}}$$ now its suffice to prove $\lim_{n \rightarrow \infty}\frac{ln n}{n}=0$  so $$n \rightarrow \infty \space \space ln n < \sqrt{n}\\ 0\leq\lim_{n \rightarrow \infty}\frac{ln n}{n} \leq \lim_{n \rightarrow \infty}\frac{\sqrt{n}}{n} \rightarrow 0 \\ \lim_{n \rightarrow \infty}e^{\frac{ln n}{n}}=e^0=1$$
A: You can also have $n>\frac{n(n+1)}{2}\epsilon_n^2$, then $\sqrt{\frac{2}{n+1}}>\epsilon_n>0$ which in the limit $n\to\infty$ implies that $\epsilon_n\to0$.
